Question

sin 2B = 2 sin B is true when B is equal to​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\rm :\longmapsto\:sin2B = 2sinB$

can be rewritten as

$\rm :\longmapsto\:sin2B - 2sinB = 0$

We know,

$\red{ \boxed{ \sf{ \:sin2x = 2sinx \: cosx}}}$

So, using this, we get

$\rm :\longmapsto\:2 \: sinB \: cosB \: - \: sinB \: = \: 0$

$\rm :\longmapsto\:sinB(2cosB - 1) = 0$

$\bf :\longmapsto\:sinB = 0 \: \: \: or \: \: \: cosB = \dfrac{1}{2}$

Case :- 1

$\rm :\longmapsto\:sinB = 0$

$\bf\implies \:B \: = \: n\pi \: \: \forall \: n \: \in \: Z$

Case :- 2

$\rm :\longmapsto\:cosB = \dfrac{1}{2}$

$\rm :\longmapsto\:cosB = cos\dfrac{\pi}{3}$

$\bf\implies \:B = 2n\pi \: \pm \: \dfrac{\pi}{3} \: \: \forall \: n \: \in \: Z$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \: \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \: \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \: \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y\: \forall \: n \: \in \: Z \\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \: \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \: \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$